The number of neutrons accompanyng the formation of _(54)^(139)Xe and _(38)^(139)Xe from the absorption of a slow neutron by _(92)^(235)U followed by nuclear fision is:

The number of neutrons accompanyng the formation of _(54)^(139)Xe and

1.	The number of neutrons accompanyng the formation of _(54)^(139)Xe and _(38)^(139)Xe from the absorption of a slow neutron by _(92)^(235)U followed by nuclear fision is:
Answer» Solution :SINCE the charge in mass is only due to the emmission of `alpha`-particle, we have Number of `alpha`-particle emitted `=(234-206)/(4)=7` Now the associated decrease in atomic number would be `14 (=2xx7)` and thus the atomic number of the daughter atom would be `76(=90-14)`. But the actual atomic number of lead is `82 E`. the atomic number is six more than EXPECTED. This is because of the emission `beta`-particle. Since there is an INCREASE of one in atomic number due to the emission of one `beta`-particle, we have Number of `beta`-particles emitted `=(82-76)/(1)=6` Hence, number of `alpha`-particles emitted `=7` ltbr,. Number of `beta`-particles emittes `=6` ANSWER is `6+7=13`